Geometry A. [3] For her daughter s th birthday, Ingrid decides to bake a dodecagon pie in celebration. Unfortunately, the store does not sell dodecagon shaped pie pans, so Ingrid bakes a circular pie first and then trims off the sides in a way such that she gets the largest regular dodecagon possible. If the original pie was 8 inches in diameter, the area of pie that she has to trim off can be represented in square inches as aπ b where a, b are integers. What is a + b? It is clear that the regular dodecagon that she ended up with has radius 4. Then the area of each slice of this pie is an isosceles triangle with interior angle 30 and two legs of side length 4. The area of this slice is (4)(4) sin 30 = 4. So the area of the entire dodecagon is 4 = 48. The original area of the pie was 4 π = 6π which means Ingrid had to cut off 6π 48 square inches of pie. 6 + 48 = 64.. [3] Terry the Tiger lives on a cube-shaped world with edge length. Thus he walks on the outer surface. He is tied, with a leash of length, to a post located at the center of one of the faces of the cube. The surface area of the region that Terry can roam on the cube can pπ be represented as q + a b + c for integers a, b, c, p, q where no integer square greater than divides b, p and q are coprime, and q > 0. What is p + q + a + b + c? (Terry can be at a location if the shortest distance along the surface of the cube between that point and the post is less than or equal to.) Figure : Figures and First note that Terry cannot get to the face opposite the face that the post is on. So if we unfold the remaining 5 faces in a cross shape as in the figure, we can see that the region that Terry can get to is the intersection of this cross with a circle of radius centered at P, where the post is. We divide up the area into 4 sectors and 8 triangles, as done in the figure. We can see that each sector s central angle is 60 and each triangle has a base of 3 and a height of. Therefore the total area is: ( 3 ) 8π 4 π +8 = +4 3 4 6 3
So our answer is 8 + 3 + 4 + 3 4 = 4. 3. [4] Cyclic quadrilateral ABCD satisfies ADC = BAD = 80 and BC = CD. Let the angle bisector of BCD meet AD at P. What is the measure, in degrees, of BP D? Let S be a point closer to D than C such that ASC is an equilateral triangle and let T be the intersection of the angle bisector of ACD and the circumcircle ω of ABCD. Note that A is the center of the circumcircle of CDS, so CSD = CAD = 0 = BCD BCA = ACP and SCD = SAC CSD = 0 = CAD, so AP C CDS. Since AC = CS, AP C CDS. Now, we have BAD = 40 = ADC = T CD since CAD is isosceles. Furthermore, since ABCT is a trapezoid as ABC = 00 = BCA + ACD = BCT, so AB = CT. Since AP C CDS, AP = CD. Thus, AP B CDT, so AP B = CDT = ACD + ADC = 0, so BP D = 60. 4. [4] Find the largest r such that 4 balls each of radius r can be packed into a regular tetrahedron with side length. In a packing, each ball lies outside every other ball, and every ball lies a + b inside the boundaries of the tetrahedron. If rcan be expressed in the form c where a, b, c are integers such that gcd(b, c) =, what is a + b + c? Let the radius be r. It is obvious that the largest r is achieved when the four balls are all tangent to three faces of the regular tetrahedron, one ball at a corner. A direct calculation shows that the distance from the center of a ball to that vertex of tetrahedron is 3r, and the distance from the center of a ball to the center of tetrahedron is 3 r, and therefore the distance from a vertex to the center of the tetrahedron is (3 + 3 )r. So the side length of the tetrahedron is ( 6 + )r and therefore r =, and our answer is 5. Author: Xiaoyu Xu = 6 6+ 0
5. [5] Let P, A, B, C be points on circle O such that C does not lie on arc BAP, P A =, P B = 56, P C = 35 and m BP C = 60. Now choose point D on the circle such that C does not lie on arc BDP and BD = 39. What is AD? 6 First, using the Law of Cosines on BP C, we get that BC = 49. Now m BOC = 60 = 0, there is a point A on the circle such that A BC is an equilateral triangle. Then since P B = 56 > 49, we know that P is on minor arc Ȃ C. If we apply Ptolemy s theorem on the cyclic quadrilateral P A BC, we get that: P A BC + P C A B = P B A C Plugging in numbers gives P A = and since A lies on the arc BAP, we know that A and A are the same point and hence ABC is equilateral. Now using Law of Cosines again, we get that AD + BD AD BD cos 0 = AB and so AD = 6. 6. [6] Triangle ABC is inscribed in a unit circle ω. Let H be its orthocenter and D be the foot of the perpendicular from A to BC. Let XY Z be the triangle formed by drawing the tangents to ω at A, B, C. If AH = HD and the side lengths of XY Z form an arithmetic sequence, the area of ABC can be expressed in the form p for relatively prime positive integers p, q. q What is p + q? Answer: Let O be the center of ω and extend AD to P and AO to Q. Let K be the intersection of AO and BC. Then, we have HBC = CAD = CBP, so HD = DP and since P Q BC, AK = AD = and OK =. KQ DP 3 WLOG X is opposite A, Y opposite B, and Z opposite C. Let M be the midpoint of Y Z and N the midpoint of minor arc Y Z of (XY Z). We claim that X, K, M are collinear. Let T be the intersection of BC and a line l through X parallel to Y Z. Now, from the law of sines on triangle pairs XT C, XT B and OKC, OKB, we have: T C T B = sin C sin 90 + X sin X+Z sin 90 X = sin Z sin Y = sin COK sin OKC sin BOK sin OKB = KC KB 3
So (B, C; K, T ) =, and since XT Y Z, if M is the intersection of XK and Y Z, then Y M = (since XT intersects Y Z at infinity) and M = M. Thus, X, K, M are collinear, as ZM desired. Now, since AO is the angle bisector of Y XZ, MY N = CXO, and since OC XY and NM Y Z, we have MY N CXO. Furthermore, NM OK since NM, OK Y Z, so NXM OXK. Let α = NY. XN Angle chasing, we have NY O = NOY, so NO = NY and = + α. XO XO Then, 3 OK = OC = NM +α = OK +α, so = 3 and α =. This means that α α α Y Z = Y M = α XC = XC, so XY + XZ = 3Y Z. Then, since the side lengths of XY Z form an arithmetic sequence, it must be that WLOG Y Z = 3k, XZ = 4k, XY = 5k for some k. The area of XY Z is then [XY Z] = 3k 4k = 6k, as XY Z is a right triangle. The inradius r of XY Z then satisfies: 6k = [XY Z] = (3k + 4k + 5k)r = 6k r r = k Since r =, we must have k =. Then, it is easy to calculate the areas of BOC, COA, AOB using similar triangles. We have: [BOC] = ( OB OX ) 3 [OBXC] = ( 3) = 3 3 + 3 0 [COA] = 5 [AOB] = Then, summing, [ABC] = + 3 + = = 6, so p = 6, q = 5 and p + q =. 0 5 0 5 7. [7] Triangle ABC has AB = AC = 0 and BC = 5. Let D be the point in ABC such that ADB BDC. Let l be a line through A and let BD and CD intersect l at P and Q, respectively. Let the circumcircles of BDQ and CDP intersect at X. The area of the locus of X as l varies can be expressed in the form p π for positive coprime integers p and q. q What is p + q? Let BQ and CP intersect at E, and by Pascal s converse on the circumcircle of BCD, hexagon BBDCCE, and line l, BCDE is concyclic. Let O be the center of the circumcircle of BCD, and note that O is also diametrically opposite A in the circumcircle of ABC. X is then the Miquel point of cyclic quadrilateral BCDE. By some of Yufei s lemmas (see his quadrilateral handouts), we know that X lies on l and that OX l. Thus, as l varies, X varies along the circumcircle of ABC, which has circumradius abc = 0 0 5 4A 55 = 80 55 and thus has area 6400 = 80 55. The desired answer is then 9. 4 5 5 4
Diagram of the case when AP = AQ 8. [8] The incircle of acute triangle ABC touches BC, AC, and AB at points D, E, and F, respectively. Let P be the second intersection of line AD and the incircle. The line through P tangent to the incircle intersects AB and AC at points M and N, respectively. Given that a AB = 8, AC = 0, and AN = 4, let AM = where a and b are positive coprime integers. b What is a + b? Lemma. M C, N B, P D and EF form a harmonic pencil of lines. Proof. By Newton s Theorem, M C, N B, P D and EF have a common point X. So AM BXCN is a complete quadrilateral, and according to the property of complete quadrilaterals, the diagonals AX and M N harmonically divide the third diagonal BC. Let G be the intersection of lines M N and BC, then CDBG is a harmonic range of points. Similarly, by Ceva s Theorem, AD, BE and CF has a common point Y, and similar to the discussion above, AF BY CE is a complete quadrilateral and the diagonals AY and EF harmonically divide the third diagonal BC. Let G be the intersection of lines EF and BC, then CDBG is a harmonic range of points. Therefore G = G, and G, E, F and P are collinear. So the fact that CDBG is a harmonic range of points implies that P C, P D, P B and P G, or equivalently, M C, N B, P D and EF, form a harmonic pencil of lines. 5
Now back to the original problem. The lemma implies that AMF B is a harmonic range of points, because these points are the intersections of the harmonic pencil of lines and the line AB. So according to the property of harmonic range of points, we have the equation AM + AB = AF Similarly we have AN + AC = AE The lengths of the two tangential segments AE and AF are obviously equal. AM + AB = AN + AC And this gives AM = 40, so our answer is 49. 9 6